Answer:
[tex]a) ~f_{avg} = 6\\b) ~c = 4,2[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]f(x)=6(x-3)^2[/tex]
a) average value of the function on the interval [2,5]
[tex]f_{avg} = \displaystyle\frac{f(5)-f(2)}{5-2}\\\\= \frac{6(5-3)^2 - 6(2-3)^2}{3}\\\\=\frac{24-6}{3} = \frac{18}{3} = 6\\\\f_{avg} = 6[/tex]
b) Value of c
[tex]f_{avg} = f(c)\\6(c-3)^2 = 6\\(c-3)^2 = 1\\c-3 = \pm 1\\c-3 = 1, c-3 = -1\\c = 4, c =2[/tex]
